3
$\begingroup$

In the wikipedia article about the pushforward, it is stated that if $f: M\to N$ is smooth, then it induces a bundle map $df: TM \to TN$. It is then claimed that equivalently $f_*=df$ is a bundle map from $TM$ to the pullback bundle $f^* TN$.

Why is this equivalent? The bundles $TN$ and $f^* TN$ are clearly not the same as they are bundles over different spaces. They could be isomorphic, but is still seems strange that this would hold independently of $f$.

Edit: The full quote is

Equivalently (see bundle map), φ∗ = dφ is a bundle map from TM to the pullback bundle φ∗TN over M, which may in turn be viewed as a section of the vector bundle Hom(TM, φ∗TN) over M. The bundle map dφ is also denoted by Tφ and called the tangent map. In this way, T is a functor.

$\endgroup$
  • $\begingroup$ $TM$ and $f^*TN$ are both bundles over $M$, not bundles over different spaces. $\endgroup$ – user10354138 Jun 11 at 15:07
  • $\begingroup$ @user10354138 You are right. But whis was only a typo. As can be seen from the context of the question, I am interested in $TN$ and $f^*TN$. I corrected the question accordingly $\endgroup$ – Michael Jun 11 at 15:41
  • $\begingroup$ Please, provide a complete quote, afik, the wikipedia article contains no such claim. (It says, however, correctly, that "Then, applying the differential pointwise to $X$ yields the pushforward $\phi_*X$, which is a vector field along $\phi$, i.e., a section of $\phi^*TN$ over $M$." The word "yields" is not the same as "is". $\endgroup$ – Moishe Kohan Jun 11 at 15:55
  • $\begingroup$ @MoisheKohan: I added the quote $\endgroup$ – Michael Jun 12 at 11:52
  • $\begingroup$ OK, then it is just sloppyness in the article; keep in mind that different parts could have been written by different people. The correct statement is "Then, applying the differential pointwise ..." $\endgroup$ – Moishe Kohan Jun 12 at 13:52
4
$\begingroup$

Recall the definition of the pull-back bundle (in the context you are interested in): If $f: M\to N$ is a smooth map, $df: TM\to TN$ is the differential of $f$, then (as a topological space) $$ f^*(TN)=\{(x,\xi): x\in M, \xi\in TN, \xi\in T_{f(x)}N\}\subset M\times TN. $$ The bundle structure on $f^*(TN)$ is given by the projection $$ \pi: (x,\xi)\mapsto x. $$ Now, $df$ defines a bundle map $TM\to f^*(TN)$ which I will denote $Df$: $$ Df: (x,\eta)\mapsto (x, df_x(\eta)), x\in M, \eta\in T_xM. $$ I will leave it to you to check that $Df$ is indeed a morphism of vector bundles $$ Df: TM\to f^*(TN). $$ By abuse of notation, one frequently denotes the morphism $Df$ simply $df$ since the"interesting" parts of these morphisms are the same.

Given a vector field $X\in {\mathfrak X}(M)$, the push-forward $f_*(X)$ is the section $Y$ of $f^*(TN)$ which is given by the formula $$ Y_x= Df(X_x). $$ That's all what there is to it. The wikipedia article commits a minor and common abuse of notation by using the notation $df$ for $Df$.

$\endgroup$
  • $\begingroup$ Thank you, this written very clearly. Just one last thing: If $f$ is a submersion, does this mean $TN \simeq f^* TM$? $\endgroup$ – Michael Jun 14 at 9:33
  • $\begingroup$ @Michael: No, it does not: Ranks could be different. What you need is a local diffeomorphism. $\endgroup$ – Moishe Kohan Jun 14 at 12:13
  • $\begingroup$ Really? What about if $f: X\times Y\to X$ is a projection? $\endgroup$ – Michael Jun 15 at 10:40
  • $\begingroup$ Yes, really. As I said, you need the equality of ranks of $TN$ and $f^*TM$: It is clearly a necessary condition for isomorphism of these bundles. Note that the bundles $f^*TM$ and $TM$ have equal rank. But rank of the tangent bundle equals the dimension of the manifold. Hence, you are asking for $dim(M)=dim(N)$. But if you have a submersion between manifolds of equal dimension, it has to be a local diffeomorphism. $\endgroup$ – Moishe Kohan Jun 15 at 13:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.